Math Problem Statement
Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set
x3 = t
and solve for x1 and x2.)
x1
−
3x3
=
−5
3x1
+
x2
−
2x3
=
1
2x1
+
2x2
+
x3
=
1
(x1, x2, x3) = complete it and show correct steps
Solution
To solve this system of equations, let's set up the augmented matrix and then use Gauss-Jordan elimination to put it into reduced row echelon form (RREF):
The system of equations is:
Step 1: Set Up the Augmented Matrix
We can rewrite this system as an augmented matrix:
1 & 0 & -3 & | & -5 \\ 3 & 1 & -2 & | & 1 \\ 2 & 2 & 1 & | & 1 \\ \end{bmatrix}$$ ### Step 2: Make Leading Entry in Row 1 a 1 Row 1 already has a leading 1 in the $$x_1$$ column, so we don’t need to make any changes. ### Step 3: Eliminate $$x_1$$ in Rows 2 and 3 - **Row 2**: Subtract 3 times Row 1 from Row 2 to make the coefficient of $$x_1$$ in Row 2 zero. $$\text{Row 2} = \text{Row 2} - 3 \times \text{Row 1} \Rightarrow [0, 1, 7, |, 16]$$ - **Row 3**: Subtract 2 times Row 1 from Row 3 to make the coefficient of $$x_1$$ in Row 3 zero. $$\text{Row 3} = \text{Row 3} - 2 \times \text{Row 1} \Rightarrow [0, 2, 7, |, 11]$$ After these operations, the matrix is: $$\begin{bmatrix} 1 & 0 & -3 & | & -5 \\ 0 & 1 & 7 & | & 16 \\ 0 & 2 & 7 & | & 11 \\ \end{bmatrix}$$ ### Step 4: Make Leading Entry in Row 2 a 1 Row 2 already has a leading 1 in the $$x_2$$ column, so we move to eliminate $$x_2$$ from Row 3. ### Step 5: Eliminate $$x_2$$ in Row 3 - **Row 3**: Subtract 2 times Row 2 from Row 3 to make the coefficient of $$x_2$$ in Row 3 zero. $$\text{Row 3} = \text{Row 3} - 2 \times \text{Row 2} \Rightarrow [0, 0, -7, |, -21]$$ The matrix now is: $$\begin{bmatrix} 1 & 0 & -3 & | & -5 \\ 0 & 1 & 7 & | & 16 \\ 0 & 0 & -7 & | & -21 \\ \end{bmatrix}$$ ### Step 6: Make Leading Entry in Row 3 a 1 Divide Row 3 by $$-7$$ to make the leading coefficient in Row 3 equal to 1: $$\text{Row 3} = \frac{\text{Row 3}}{-7} \Rightarrow [0, 0, 1, |, 3]$$ The matrix now is: $$\begin{bmatrix} 1 & 0 & -3 & | & -5 \\ 0 & 1 & 7 & | & 16 \\ 0 & 0 & 1 & | & 3 \\ \end{bmatrix}$$ ### Step 7: Back-Substitute to Eliminate $$x_3$$ in Rows 1 and 2 - **Row 1**: Add 3 times Row 3 to Row 1 to make the coefficient of $$x_3$$ in Row 1 zero. $$\text{Row 1} = \text{Row 1} + 3 \times \text{Row 3} \Rightarrow [1, 0, 0, |, 4]$$ - **Row 2**: Subtract 7 times Row 3 from Row 2 to make the coefficient of $$x_3$$ in Row 2 zero. $$\text{Row 2} = \text{Row 2} - 7 \times \text{Row 3} \Rightarrow [0, 1, 0, |, -5]$$ The final matrix in RREF is: $$\begin{bmatrix} 1 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & | & -5 \\ 0 & 0 & 1 & | & 3 \\ \end{bmatrix}$$ ### Solution From this matrix, we can read off the solution: $$x_1 = 4, \quad x_2 = -5, \quad x_3 = 3$$ So the solution to the system is: $$(x_1, x_2, x_3) = (4, -5, 3)$$ Would you like more details or have any questions? --- 1. How does Gaussian elimination differ from Gauss-Jordan elimination? 2. What are the steps involved in back-substitution? 3. How do you recognize if a system has no solution during elimination? 4. What are the implications of infinite solutions in a system of equations? 5. How is the solution verified once found? **Tip:** When solving systems with matrices, always double-check row operations to avoid calculation errors!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Systems of Linear Equations
Gaussian Elimination
Gauss-Jordan Elimination
Back-Substitution
Matrix Operations
Formulas
Gaussian Elimination
Gauss-Jordan Elimination
Back-Substitution
Theorems
Row Echelon Form
Reduced Row Echelon Form (RREF)
Linear System Consistency
Suitable Grade Level
Grades 9-12
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